What is (\partial)/(\partial x)(sqrt(4x^2+5y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(4x^2+5y^2)) is (4x)/(sqrt(4x^2+5y^2))

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(\partial)/(\partial x)(sqrt(4x^2+5y^2))

\frac{\partial}{\partial x} (4 x^{2} + 5 y^{2})

\frac{4 x}{4 x ^{2} + 5 y ^{2}}

Solution steps

\frac{\partial}{\partial x} (4 x^{2} + 5 y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 4 x ^{2} + 5 y ^{2}} \frac{\partial}{\partial x} (4 x^{2} + 5 y^{2})

= \frac{1}{2 4 x ^{2} + 5 y ^{2}} \frac{\partial}{\partial x} (4 x^{2} + 5 y^{2})

\frac{\partial}{\partial x} (4 x^{2} + 5 y^{2}) = 8 x

= \frac{1}{2 4 x ^{2} + 5 y ^{2}} \cdot 8 x

Simplify

\frac{1}{2 4 x ^{2} + 5 y ^{2}} \cdot 8 x : \frac{4 x}{4 x ^{2} + 5 y ^{2}}

= \frac{4 x}{4 x ^{2} + 5 y ^{2}}

\frac{d}{d x} (ln (\frac{1}{x} + ln (\frac{1}{x} + ln (\frac{1}{x}))))

n = 1 \sum \infty n e^{- n^{2}}

\int \frac{6 x}{( x ^{2} + 5 ) ^{2}} d x

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 10, y^{^{'}} (0) = 0

\int_{0}^{\frac{π}{4}} sin (2 x) d x

What is (\partial)/(\partial x)(sqrt(4x^2+5y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(4x^2+5y^2)) is (4x)/(sqrt(4x^2+5y^2))